重难点13 极化恒等式与等和(高)线定理【四大题型】
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1 !"#$
« 1&2024·¬®¯·HG);a △𝐴𝐵𝐶8;D 3𝐵𝐶E/8Y;E;F 3+°𝐴𝐷/JH!±Y;
𝐵𝐴⋅𝐶𝐴= 7;𝐵𝐸⋅𝐶𝐸= 2;t𝐵𝐹⋅𝐶𝐹= & )
Ag−2
Bg−1
Cg1
Dg2
¥" 1-1&23-24 Hy·²N³´·µ¶);BC0DE 3·~ 1 /¸ O /J¹·;𝐵𝐹= 2𝐹𝑂;
t𝐹𝐷⋅𝐹𝐸= & )
Ag−
3
4
Bg−
8
9
Cg−
1
4
Dg−
4
9
¥" 1-2&2024 H·º»·¼½);a45DEF ABCD 8;O ~ BD /8Y;¾ OA¿3;OC
¿5.𝐴𝐵⋅𝐴𝐷= À7;t𝐵𝐶⋅𝐷𝐶/$3
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¥" 1-3&23-24 Á·ÃÄSÅ·ÆÇ);a4CDEF ABCD 8;AB¿1;AD¿2;Y E;
F;G;H ±È3 AB;BC;CD;AD Ey/8Y;t𝐸𝐹⋅𝐹𝐺+ 𝐺𝐻⋅𝐻𝐸!p
g
2 !"#%$&'()
« 2&2024 H·ÉÊ·¼½)·~2/¸𝑂yBHY𝐴0𝐵0𝐶ËÌ𝑂𝐴+ 𝐴𝐵+ 𝐴𝐶= 0;Y𝑃3¸
Y;t𝑃𝐴⋅𝑃𝑂+ 𝑃𝐵⋅𝑃𝐶/Í$'(~& )
Ag[−4,14)
Bg[0,4)
Cg[4,14]
Dg[4,16]
¥" 2-1&23-24 ·º»Äηµ8)©HIF𝐴𝐵𝐶/ES~ 3;Y𝐷aE𝐴𝐵y;¾𝐵𝐷= 2𝐷𝐴;HI
F𝐴𝐵𝐶/Ïи/¹Ñ𝑀𝑁jY𝐷;Y𝑃~E𝐵𝐶y/ÒY; Ñ𝑀𝑁/ST%Ó¢;𝑃𝑀⋅𝑃𝑁/Í$'(3
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Ag[−1,5]
Bg[−1,7]
Cg[0,2]
Dg[1,5]
¥" 2-2&2024·9Ô·GÕÖ×)Øe △𝑂𝐴𝐵/5>~1,𝐴𝐵= 2;ÒY𝑃,𝑄a+°𝐴𝐵yÙÒ;¾|𝑃𝑄|
= 1;t𝑂𝑃⋅𝑂𝑄/%Ú$~
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¥" 2-3&23-24 Hy·yÛÜÝ·Þ°¼½)a5>~ 2 /4CDEF8𝐴𝐵𝐶𝐷8;∠𝐷𝐴𝐵=
π
6;Y P
3 AD ßa+y/ÒY;t𝑃𝐵
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« 3&2024·Dàá·GÕÖ×);a4CDEF𝐴𝐵𝐶𝐷8;𝐵𝐸=
2
3𝐵𝐶;𝐷𝐹=
3
4𝐷𝐸;𝐴𝐹= 𝜆𝐴𝐵+𝜇
𝐴𝐷;t𝜆+ 𝜇= &ââ)
Ag
3
2
Bg−
1
12
Cg
1
12
Dg0
¥" 3-1&2023·ãäå·GÕÖ×)a △𝐴𝐵𝐶8𝐵𝐸=
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2𝐸𝐶,𝐵𝐹=
1
2 𝐵𝐴+ 𝐵𝐶;Y𝑃~𝐴𝐸1𝐵𝐹/æY;
𝐴𝑃= 𝜆𝐴𝐵+𝜇𝐴𝐶;t𝜆+ 𝜇= & )
Ag0
Bg
1
4
Cg
1
2
Dg
3
4
¥" 3-2&23-24 y·º»·µ¶)a4CDEF𝐴𝐵𝐶𝐷8;𝐸~𝐵𝐶/8Y;𝐹a+°𝐷𝐶y;¾
𝐶𝐹= 2𝐷𝐹g𝐴𝐶= 𝜆𝐴𝐸+𝜇𝐴𝐹;𝜆,𝜇ç~è
;t𝜆+ 𝜇/$~
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¥" 3-3&23-24 y·º»»·µ¶);aéF𝐴𝐵𝐶𝐷8;𝑀;𝑁±È~+°𝐵𝐶;𝐶𝐷/8Y;
𝑀𝑁= 𝜆1𝐴𝑀+ 𝜆2𝐵𝑁;𝜆1,𝜆2 ∈R;t𝜆1 + 𝜆2/$~
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𝐴𝐵+𝑦𝐴𝐶;t2𝑥+ 2𝑦/%K$~& )
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8
3
Bg2
Cg
4
3
Dg1
¥" 4-1&23-24 Hy·ãäå·µ8); △𝐵𝐶𝐷1 △𝐴𝐵𝐶/5>Uª~ 2;Y P 3íî𝐴𝐵𝐶𝐷
lY&ïEð);¾𝐴𝑃= 𝜆𝐴𝐵+𝜇𝐴𝐶(𝜆,𝜇∈𝑅);t𝜆+ 𝜇/Í$'(3& )
Ag[0,1]
Bg[0,2]
Cg[0,3]
Dg[0,4]
¥" 4-2&23-24 ·²Nñ·Þ°¼½)a △𝐴𝐵𝐶8;M ~ BC EylY;N ~+° AM y
lY;𝐴𝑁= 𝜆𝐴𝐵+𝜇𝐴𝐶&𝜆;𝜇∈𝑅);t𝜆+ 𝜇/Í$'(3
.
¥" 4-3&23-24 ·òóôõ·µ¶)Øe𝑂~ △𝐴𝐵𝐶Y;¾4𝑂𝐴+8𝑂𝐵+5𝑂𝐶= 0;Y𝑀a △𝑂𝐵𝐶
&qïEð);𝐴𝑀= 𝜆𝐴𝐵+𝜇𝐴𝐶;t𝜆+ 𝜇/Í$'(3
g
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1g&2024·Dà÷ø·HG);a△𝐴𝐵𝐶8;𝐴𝐹= 𝐵𝐹= 6,𝐸𝐹= 5;t𝐸𝐴⋅𝐸𝐵= & )
Ag−11
Bg−13
Cg−15
Dg15
2g&2024·ùóóú·G)a △𝐴𝐵𝐶8;Y𝐷3+°𝐴𝐶yY;Y𝑃3+°𝐵𝐷yY;¾𝐶𝐷= 𝐷𝐴;𝐴𝑃=
2
3𝐴𝐵+𝜆𝐴𝐶;t𝜆= & )
Ag
1
6
Bg
1
3
Cg
2
3
Dg
5
6
3g&2024 H·ÉÊ·¼½)a △𝐴𝐵𝐶8;𝐷3𝐵𝐶Ey/8Y;¾𝐴𝐸=
1
3𝐴𝐷;𝐴𝐹= 2𝐴𝐸;𝐴𝐵⋅𝐴𝐶= 6;
𝐹𝐵⋅𝐹𝐶= −2;t𝐸𝐵⋅𝐸𝐶 = & )
Ag−1
Bg2
Cg−
1
2
Dg1
4g&2024·ùóûõ·HG)a △𝐴𝐵𝐶8;𝐸aE𝐵𝐶y;¾𝐸𝐶= 3𝐵𝐸,𝐷3E𝐴𝐵ylY;𝐴𝐸1𝐶𝐷æpY
𝑃;𝐶𝑃= 𝑥𝐶𝐴+𝑦𝐶𝐵;t3𝑥+ 4𝑦= & )
Ag
3
4
Bg−
3
4
Cg3
Dg-3
5g&23-24 H·ÃÄSÅ·Þ°¼½)67/
7>LM|}~
M67~E/4CDEF/“*
nI+”1“nI+”4o/D±U;
ß};𝑎⋅𝑏=
1
4 |𝐴𝐷|
2−|𝐵𝐶|
2 ;~ !". Ø
ea △𝐴𝐵𝐶8;𝑀3𝐵𝐶8Y;𝐴𝑀= 3;𝐵𝐶= 10;t𝐴𝐵⋅𝐴𝐶= & )
Ag−16
Bg16
Cg−8
Dg8
6g&2024·ÉÊ·GÕÖ×);a △𝐴𝐵𝐶8;𝐴𝑁= 𝑡𝑁𝐶(𝑡> 0);𝐵𝑃= 𝜆𝑃𝑁(𝜆> 0);𝐴𝑃=
3
4𝐴𝐶−
1
4𝐵𝐶;
t𝜆+ 𝑡/$~& )
Ag7
Bg6
Cg5
Dg4
7g&23-24 Hy·êÝüý·µ¶)Øe©oF ABCD /ES~ 2;MN 3þ/ÿ¸/¹Ñ;Y P ~©
oFD¹Ey/ÒY; Ñ MN /ST%K¢;𝑃𝑀⋅𝑃𝑁/Í$'(3& )
Ag[0;1]
Bg[0, 2]
Cg[1;2]
Dg[−1,1]
8g&2024·ãäå·HG)nĀ3
Ā/9:ā±;ĂăĄapą!
*!
/Ʊ
ć8;a
Ĉy;
Ā3
ĉĊ/Òċ.;a!E △𝐴𝐵𝐶8;𝐴𝐵= 2;MH¹E~·6ÏcH
¸;𝑀3H¸Čy/ÒY;𝐵𝑀= 𝜆𝐴𝐵+𝜇𝐴𝐶;t𝜆+ 𝜇/%K$~& )
Ag
1
2
Bg 3
3
Cg1
Dg
3
2
Á0č
9g&23-24 ·º»ÄĎ·µ8)a △𝐴𝐵𝐶8;Y𝐷3+°𝐵𝐶ylY;Y𝑀3+°𝐴𝐷/8Y;
a𝜆,𝜇∈R𝐵𝑀= 𝜆𝐴𝐵+𝜇𝐴𝐶;t𝜆,𝜇/Í$Lď3& )
Ag𝜆= −
3
5,𝜇=
1
10
Bg𝜆= 1,𝜇= −
3
2
Cg𝜆= −
9
10,𝜇=
2
5
Dg𝜆= −
7
10,𝜇=
3
5
10g&23-24 ·Dàá·Þ°¼½);©oF𝐴𝐵𝐶𝐷8;𝐸~𝐴𝐵8Y;𝑀~+°𝐴𝐷y/ÒY;
𝐵𝑀= 𝜆𝐵𝐸+𝜇𝐵𝐷;t𝜆+ 𝜇/$LM3& )
Ag
3
2
Bg
1
2
Cg1
Dg2
11g&23-24 ·ùóóú·Þ°¼½)&č);aDEF𝐴𝐵𝐶𝐷8;∠𝐵= 60∘;𝐴𝐵= 3;𝐵𝐶= 6;
¾𝐴𝐷= 𝜆𝐵𝐶(𝜆∈R);𝐴𝐷⋅𝐴𝐵= −
3
2;t& )
Ag𝐴𝐵·𝐵𝐶= 9
Bgè
𝜆/$~
1
6
CgDEF𝐴𝐵𝐶𝐷3ĐF
Dg𝑀;𝑁3+°𝐵𝐶y/ÒY;¾|𝑀𝑁| = 1;t𝐷𝑀⋅𝐷𝑁/
%Ú$~
13
2
H0đĒ
12g&2024·ē·ÁG)a!ĔĐF𝐴𝐵𝐶𝐷8;𝐴𝐵= 2𝐷𝐶;Y𝐸3+°𝐵𝐶/8Y;𝐴𝐸= 𝜆𝐴𝐵+𝜇𝐴𝐷;t
𝜆+ 𝜇=
.
13g&23-24 ·ĕĖºKÔ·µ¶);a △𝐴𝐵𝐶8;𝐷3𝐵𝐶/8Y;𝐸;𝐹3𝐴𝐷y/JH!±Y
𝐵𝐴⋅𝐶𝐴= 5;𝐵𝐹⋅𝐶𝐹= −2;t𝐵𝐸⋅𝐶𝐸/$3
.
14g&23-24 H·òÝøº·Þ°¼½)a5>~2/4CDEF𝐴𝐵𝐶𝐷8;Y𝑃~+𝐴𝐷y/ÒY;t𝑃𝐵⋅𝑃𝐶
+ 𝐵𝐶2/%Ú$3
g
D0<ė
15g&23-24 ·ĘęĚě·Þ°¼½);a4CDEF𝐴𝐵𝐶𝐷8;𝐴𝐶1𝐵𝐷wæpY𝑂g𝐸3+°𝑂𝐷
/8Y;𝐴𝐸/ĜS+1𝐶𝐷æpY𝐹g
(1)𝐴𝐵;𝐴𝐷o|}𝐴𝐸W
(2)𝐴𝐹= 𝜆𝐴𝐵+𝜇𝐴𝐷;#𝜆+ 𝜇/$g
16g&23-24 ·º»»·µ8)ĝаğ
𝑎+ 𝑏
2 = 𝑎2 +2𝑎⋅𝑏+ 𝑏2; 𝑎−𝑏
2 = 𝑎2−2𝑎⋅𝑏+
𝑏2;J"wz_(𝑎+ 𝑏)2−(𝑎−𝑏)2 = 4𝑎·𝑏⇒𝑎·𝑏=
1
4[(𝑎+ 𝑏)2−(𝑎−𝑏)2] !"c“ !"”;þ
èĠġaĢBģI/Ĥ1ÂĥJ67/
7>Ħħ~“G”/ĦħgÇĨĩy5/Ī<=MÂ?
;a△ABC 8;D 3 BC /8Y;E;F 3 AD y/JH!±Yg
(1) AD¿6;BC¿4;#𝐴𝐵⋅𝐴𝐶/$W
(2)𝐴𝐵⋅𝐴𝐶= 4;𝐹𝐵⋅𝐹𝐶= −1;#𝐸𝐵⋅𝐸𝐶/$g
17g&23-24 y·īĬKĭ·µ¶)aHIF𝐴𝐵𝐶8;𝐴𝐵= 𝑎;𝐴𝐶= 𝑏;𝐵𝐸= 2𝐸𝐶;𝐷~+°𝐴𝐶yl
Y;𝐵𝐷æ𝐴𝐸p𝑂.
(1)𝐶𝐷= 2𝐷𝐴.
①𝑎;𝑏|}𝐴𝐸W
②𝐴𝑂= 𝜆𝐴𝐸;#𝜆/$W
(2)𝐵𝑂= 𝑥𝐵𝐴+𝑦𝐵𝐶;#
1
2𝑥+
1
3𝑦+1/%Ú$.
18g&23-24 ·ÃÄĮø·µ¶);ØeDEF𝐴𝐵𝐷𝐸~4CDEF;Y𝐶a𝐴𝐵ĜS+y;Y𝑀a+
°𝐴𝐷y;¾𝐴𝐵=
1
2𝐵𝐶,𝐴𝑀=
1
3𝐴𝐷;s𝐴𝐵= 𝑎,𝐴𝐸= 𝑏.
(1)67𝑎;𝑏|}𝐶𝐷W
(2)+°𝐶𝑀yaÒY𝑃;¾𝐴𝑃= 𝑚𝑎+𝑛𝑏(𝑚,𝑛∈R);#𝑛2 +𝑚𝑛/%K$.
19g&23-24 ·òÝį·Þ°¼½)ĝĞMÂİı;<=IJ
eij①(𝑎+ 𝑏)
2 = 𝑎2 +2𝑎⋅𝑏+
𝑏
2W②(𝑎−𝑏)
2 = 𝑎2−2𝑎⋅𝑏+ 𝑏
2.X①-②_(𝑎+ 𝑏)
2−(𝑎−𝑏)
2 = 4𝑎⋅𝑏⇔𝑎⋅𝑏= (𝑎+𝑏)
2−(𝑎−𝑏)
2
4
;%Ĵ]
`/"ĵ~“ !"”;þèĠġĢBģIĤ1/ĶķÂĥJ67/
7>~“G”/Ħħ.ß
}/DEF𝐴𝐵𝐶𝐷8;𝐵𝐷= 8,𝐴𝐵⋅𝐴𝐷= 48;𝐸~𝐵𝐷8Y.
(1)cos∠𝐵𝐴𝐷=
12
13;# △𝐴𝐵𝐷/5>W
(2)2𝐴𝐸= 𝐸𝐶;#𝐶𝐵⋅𝐶𝐷/$W
(3)𝑃~45𝐴𝐵𝐶𝐷Y;#𝑃𝐴⋅𝑃𝐵+ 𝑃𝐷/%Ú$.